If \(a+\displaystyle\frac{1}{a}+2=0\), then find the value of \(a^{49}-\displaystyle\frac{1}{a^{48}}\).
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(-2\)
Multiplying the condition by \(a\) turns it into a perfect square, which fixes \(a\) exactly. Then an odd power of \(-1\) is \(-1\) and an even power is \(1\).
\(a^2+1+2a=0\Rightarrow(a+1)^2=0\), so \(a=-1\)
\(a^{49}=(-1)^{49}=-1\) (odd power)
\(a^{48}=(-1)^{48}=1\) (even power), so \(\displaystyle\frac{1}{a^{48}}=1\)
\(a^{49}-\displaystyle\frac{1}{a^{48}}=-1-1=-2\)